# How find angle to horizontal of hanging string

Could anyone point out my error please, may answer disagrees with that from a book.

A light elastic string, of unstretched length a and modulus of elasticity W, is fixed at one end to a point on the ceiling of a room. The other end of the string is attached to a particle of weight W. A horizontal force P is applied to the particle and in equilibrium it is found that the string is stretched to three times its natural length. Calculate the angle that the string makes with the horizontal.

I can use the standard formula for the extension x of an elastic string of natural length a due to a mass m hanging on it obtained by equating the force of gravity and Hooke's law - which gives:

$x=\frac{mga}{\lambda}$

So in my case mg = W and $\lambda=W$ and so I have $x=a$

So the string has a length of 2a when the mass m is hung on it.

Now the particle if moved horizontally by the force P until the string has length 3a. So I have a triangle of vertical side 2a and hypotenuse 3a which gives:

$sin\, \theta=\frac{2a}{3a}=\frac{2}{3}$ and so $\theta=41.8^\circ$

But the book answer is 30 degrees.

I would obtain the right answer if I read "it is found that the string is stretched to three times its natural length" as "the string is now found to have an extension of three times it's natural length" so that it's new length would be 4a leading to $sin\,\theta=\frac{2}{4}=\frac{1}{2}$.

Thanks, Mitch.

The stretching for the string is only $\Delta L=2a$. The force along the string is $T=\lambda\Delta L/L=\lambda 2a/a=2W$ On the vertical direction the particle's weight must be equal to the vertical component of $T$:
$mg=2W\sin\alpha$ or $W=2W\sin\alpha$ leading to $\sin\alpha=1/2$
Your initial position is correct: for a force $mg$, the elongation is $a$. Following the idea, the triangle has to be constructed with the elongations, not the total lengths (length $a$ corresponds to a force $0$ because the elongation is $0$). In your triangle, the vertical side is $a$ and the hypotenuse $2a$